int rows = word1.size() + 1;

int cols = word2.size() + 1;

int ** d = (int**)malloc(rows * sizeof(int*));

for(int i = 0; i < rows; ++i){

d[i] = (int*)malloc(cols * sizeof(int));

d[i][0] = i;//sub string in word1 range [0,i) edit to ""

}

for(int j = 0; j < cols; ++j)

d[0][j] = j;//sub string in word2 range [0,j) edit to ""

for(int i = 1; i < rows; ++i) {

char ci = word1[i-1];

for(int j = 1; j < cols; ++j) {

char cj = word2[j-1];

//we will edit str1:word1[0,i) to str2:word2[0,j)

if (ci == cj) {

//if ci equal to cj, then the edit ditance of word1[0,i) to word2[0,j)

// is the same as word1[0,i-1) to word2[0,j-1)

d[i][j] = d[i-1][j-1];

} else {

//if we modify letter ci to cj, there will be 1 operation

int dEdit = d[i-1][j-1] + 1;

//if we add cj to the end of word1[0,i), then from the edit distance of

// word1[0,i) to word2[0, j -1), we can conclude follow dist

int dAdd = d[i][j-1] + 1;

//if we delete ci from word1[0,i), and we know the dist

// from word1[0,i-1) to word2[0,j), the things done:

int dDel = d[i-1][j] + 1;

//the minimum one will be the final distance for str1 to str2

int min = dEdit < dAdd ? dEdit : dAdd;

min = min < dDel ? min : dDel;

d[i][j] = min;

}

}

}

int result = d[rows - 1][cols - 1];

//delete dist;

return result;

{

vector<string> input[257];

string str;

int n = 0,p;

while(cin >> str)

{

p = 0;

int len = str.length();

for(int i = len -5 ; i < len + 5; i++)

{

if(input[i].size() > 0)

{

for(vector<string>::iterator it = input[i].begin();it != input[i].end(); it++)

if(minDistance(str,*it) <= 5)

p++;

}

}

if(p > n)

n = p;

input[len].push_back(str);

}

cout << p << endl;